Equivalent form of Uniformly Convex How to show that the following statements are equivalent.


*

*For each $\epsilon>0$, there exists a $\delta>0$ such that
$\parallel x\parallel=1=\parallel y\parallel$ and $\parallel x-y\parallel\geq\epsilon$ implies that $\parallel\frac{x+y}{2}\parallel\leq(1-\delta)$

*For each $\epsilon>0$, there exists a $\delta>0$ such that
$\parallel x\parallel\leq 1, \parallel y\parallel \leq1$ and $\parallel x-y\parallel\geq\epsilon$ implies that $\parallel\frac{x+y}{2}\parallel\leq(1-\delta)$
 A: Clearly (2) implies (1). So assume that (1) holds true and let us try to prove (2).
For any $\varepsilon >0$, let us denote by $\delta(\varepsilon)$ the positive number given by (1). Obviously, we may assume that $\delta$ is nondecreasing as a function of $\varepsilon$.
Now, let us fix $\varepsilon >0$. We are looking for some $\delta'=\delta'(\varepsilon)$ such that (2) holds true with $\delta'$. 
Let $x,y\in X$ satisfy $\Vert x\Vert, \Vert y\Vert\leq 1$ and $\Vert x-y\Vert\geq\varepsilon$. Let also $\alpha \in (0,1)$ to be chosen later.
Assume first that $\Vert x\Vert$ or $\Vert y\Vert$ is not greater than $1-\alpha$. Then $$\left\Vert\frac{x+y}2\right\Vert\leq\frac12(1+(1-\alpha))=1-\frac12\alpha\, .$$
Now, assume that $\Vert x\Vert$ and $\Vert y\Vert$ are both greater than 
$1-\alpha$. Then $u:=\frac{x}{\Vert x\Vert}$ and $v:=\frac{y}{\Vert y\Vert}$ satisfy $$\Vert u-x\Vert=\left\Vert\left(\frac1{\Vert x\Vert}-1 \right) x \right\Vert\leq \left(\frac1{1-\alpha}-1\right)\Vert x\Vert\leq \frac\alpha{1-\alpha}\, ,$$
and likewise
$$\Vert v-y\Vert\leq\frac\alpha{1-\alpha}\cdot $$ 
Since $\Vert x-y\Vert\geq\varepsilon$, it follows that $\Vert u-v\Vert\geq \varepsilon-2\frac\alpha{1-\alpha}\cdot$ Assuming $\frac{2\alpha}{1-\alpha}<\varepsilon$ and remembering that $\Vert u\Vert =1=\Vert v\Vert$, we deduce that $\Vert \frac{u+v}2\Vert\leq 1-\delta\left(\varepsilon-2\frac\alpha{1-\alpha} \right)$ and hence (using again the fact that $\Vert u-x\Vert$ and $\Vert v-y\Vert$ are not greater than $\frac\alpha{1-\alpha}$) that
\begin{eqnarray*}\left\Vert \frac{x+y}2\right\Vert&\leq&\left\Vert \frac{u+v}2\right\Vert+\frac\alpha{1-\alpha} \\&\leq & 1-\delta\left(\varepsilon-2\frac\alpha{1-\alpha} \right)+\frac\alpha{1-\alpha}\cdot 
\end{eqnarray*}
So we are able to estimate $\left\Vert\frac{x+y}2\right\Vert$ in the two cases we have considered.
Now, choose $\alpha=\alpha(\varepsilon)$ such that $$\frac\alpha{1-\alpha}<\min\left( \frac\varepsilon4, \delta(\varepsilon/2) \right)\, .$$
Observe that that since $\varepsilon-2\frac\alpha{1-\alpha}\geq\varepsilon/2$, we have $\delta\left(\varepsilon-2\frac\alpha{1-\alpha} \right)-\frac\alpha{1-\alpha}\geq\delta(\varepsilon/2)-\frac\alpha{1-\alpha}>0$. Hence, 
$$\delta'(\varepsilon)=\min\left( \frac\alpha2, \delta\left(\varepsilon-2\frac\alpha{1-\alpha} \right)-\frac\alpha{1-\alpha}\right) $$
is a positive number (depending only on $\varepsilon$); and by the above two estimates for $\left\Vert\frac{x+y}2\right\Vert$, (2) holds true for the given $\varepsilon$ with $\delta=\delta'(\varepsilon)$.
